2.019 Design of Ocean Systems
Lecture 8
Seakeeping (IV) March 4, 2011 General Response of A Floating Body in Regular Ambient Waves
z
Incident wave:
ζ3 (t) = <{ζ¯3 eiωt }
ζ6 (t) = <{ζ¯6 eiωt }
y ζ2 (t) = <{ζ̄2 eiωt }
ηI = a cos(ωt − kx)
ζ5 (t) = <{ζ̄5 eiωt }
Φtt + gΦz = 0
~n
Equation of motion:
6
X
η(t) = −Φt /g
x
ζ1 (t) = <{ζ¯1 eiωt }
ζ4 (t) = <{ζ¯4 eiωt }
[(Mj` + Aj` )ζ̈` + Bj` ζ̇` + Cj` ζ` ] = F̄Ej eiωt
(j = 1, . . . , 6)
(1)
`=1
6
X
[−ω 2 (Mj` + Aj` ) + iωBj` + Cj` ]ζ̄j = FEj
(j = 1, . . . , 6)
`=1
Mj` : 6 × 6 elements of the egeneralized mass matrix
Aj` , Bj` : 6 × 6 elements of added mass and wave damping matrices
Cj` : 6 × 6 elements of hydrostatic restoring matrix
FEj : 6 elements of the wave excitation vector
Transfer function or Response Amplitude Operator (RAO): Hj (ω) =
ζ¯j (ω)
a
(j = 1, . . . , 6)
Numerical Method for Potential-Flow Problems
Uniform free stream:
Φ = Ux
2D point source:
m
Φ =
2π
ln
m
ur = 2πr
→
u = U, v = 0, w = 0
√
x2 +
ur
m
z 2
=
2π
ln r
2D point source plus point sink:
Φ
p
p
m
2
2
(x + s) + z −
2π
ln
(x − s)2 + z 2
m
=
2π
ln
source
sink
2D doublet or dipole: source + sink,
Φ
m
= lim
ln
s→0 2π
(p
p
(x +
s)2
as s → 0 while keeping 2ms = µ.
+
z2
(x − s)2 + z 2
)
2xs
m
µ
x
√
√
=
s→0 2π
2π x2 + z 2
x2 + z 2
= lim
2D Stream plus dipole: Φ = Ux +
µ √ x
2π x2 +z 2
a=
p
µ
2πU
z
Three-dimensional point source:
Φ(~x, ξ~)
=−
Q
4πR
Q
1
=− p
4π (x − ξ)2 + (y − η)2 + (z − ζ)2
O
Three-dimensional source distribution:
Distribute sources of strength M (ξ~, t)dS, varying with space ξ and pulsating in
time t and proportional to surface area dS:
dΦ =
sources are distributed
over the surface of the body
1
~, t)G(~x, ξ~)dS
M
(ξ
4π
G(~x, ξ~): Green’s function
Green function
in unbounded fluid:
G(~x, ξ~) =
1
R
= √
1
(x−ξ)2 +(y −η)2 +(z−ζ)2
Free-surface Green function
(in finite depth H) satisfying linearized
free-surface boundary condition:
1
1
+ 0
R R
Z ∞
= +2 −
G(~x, ξ~)
=
0
(µ + ν)e−µH
cosh µ(ζ + H) cosh µ(z + H)J0 (µr)dµ
µ sinh µH − ν cosh νH
k2 − ν 2
+2πi 2
cosh k(z + H) cosh k(ζ + H)J0 (kr)
(k − ν 2 )H + ν
ν=
R=
ω2
g
p
= k tanh kH
(x − ξ)2 + (y − η)2 + (z − ζ)2 ,
R0 =
p
(x − ξ)2 + (y − η)2 + (z + 2H + ζ)2
Source Method
• Distribution sources on the body surface with unknown strengths:
M (ξ~, t) = Re{M̄ (ξ~)eiωt }
Diffraction problem: M̄D (ξ~)
Φ̄D (~x) =
1
4
π
Radiation problem: M̄j (ξ~)
Φ̄j (~x) =
1
4
π
R
S
R
S
M̄D (ξ~)G(~x, ξ~)dS
M̄j (ξ~)G(~x, ξ~)dS
• The source strength is found by requiring the velocity satisfies the boundary
condition on the body surface S
Boundary
condition
at ~x = ~xB :
Diffraction problem: R
1
1
¯
D (~x) + −M̄D (ξ~) ∂ G(~x, ξ~)dS
= −
∂Φ¯ I
−
2
M
4
π
S
∂n
∂n
Radiation problem:
¯
x)
−
1
2
Mj (~
+
R
4
1
π
−SM̄j (ξ~) ∂∂n G(~x, ξ~)dS
=
−(iω)nj
• To solve the integral equation for unknown source strengths, we apply the so-called
panel method: Subdividing the body surface into N elements with the assumption of
an uniform distribution of source strength over each element. This will leads to N
equations and N unknown source strengths:
−M̄D (~xm ) +
PN
∂Φ̄I (~
xm )
¯
α
(~
x
)
=
−
M
n=1 mn D n
∂n
m = 1, 2, · · · , N
αmn =
R
∆Sn
∂
~n )dS
G(~
x
,
ξ
m
∂n
• Once unknown source strengths on the body are found, the diffraction and radiation
potentials can be evaluated:
Φ̄D (~x) =
PN
1
M̄
(~
x
)
D
n
n=1
4π
R
~n )dS
G(~
x
,
ξ
∆Sn
• Numerical solution of the linear system of N equations: [A]{M } = {b}
Gauss elimination ∼ O(N 3 ) computational effort
Iteration solver ∼ O(N 2 ) computational effort
Convergence with error ∼ 1/N as N → ∞
For better convergence
of the solution,
discretization must be
fine where geometry
changes sharply or
near free surface.
Free surface
Flow decay
Exponentially
With depth
from free
surface
Figures removed due to copyright restrictions.
Please see Fig. 3 and Fig. 5 from Newman, J.,
and P. Sclavounos. "The Computation
of Wave Loads on Large Offshore Structures."
Figures removed due to copyright restrictions. Please see Fig. 3
and Fig. 5 from Newman, J.,and P. Sclavounos. "The Computation
of Wave Loads on Large Offshore Structures."
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2.019 Design of Ocean Systems
Spring 2011
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